A stone is thrown vertically upward with an initial velocity v0.Find the distance travelled by it in time 1.5v0/g

To find the distance traveled by a stone thrown vertically upward with an initial velocity \( v_0 \) in the time \( \frac{1.5v_0}{g} \), we can use the equations of motion under uniform acceleration. In this case, the only acceleration acting on the stone is due to gravity, which acts downwards, and its value is \( g \). Let's break this down step by step.

Understanding the Motion

When a stone is thrown upwards, it will initially rise until it reaches its highest point, where its velocity becomes zero before it starts falling back down. The time it takes to reach this highest point can be calculated and used as part of our overall analysis.

Key Equations of Motion

We can utilize the kinematic equation that relates distance, initial velocity, time, and acceleration:

• The formula for distance is given by: s = v_0 t - \frac{1}{2} g t^2

Here, \( s \) represents the distance traveled, \( v_0 \) is the initial velocity, \( t \) is the time, and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

Calculating the Distance

Now, we can substitute the time \( t = \frac{1.5v_0}{g} \) into our distance formula:

s = v_0 * (1.5v_0/g) - (1/2) * g * (1.5v_0/g)^2

Let's simplify this step by step:

Step 1: Substitute for time

s = v_0 * (1.5v_0/g)- (1/2) * g * ((1.5v_0)^2 / g^2)

Step 2: Simplify the first term

s = (1.5v_0^2 / g)

Step 3: Simplify the second term

(1/2) * g * (2.25v_0^2 / g^2) = (1/2) * (2.25v_0^2 / g)= 1.125v_0^2 / g

Step 4: Combine the terms

Now we can combine the two parts:

s = (1.5v_0^2 / g) - (1.125v_0^2 / g)= (1.5 - 1.125) * (v_0^2 / g)= 0.375 * (v_0^2 / g)

Final Result

Thus, the distance traveled by the stone in the time \( \frac{1.5v_0}{g} \) is:

This result indicates how far the stone travels upwards before its motion is significantly affected by gravity. The outcome shows that the distance is directly proportional to the square of the initial velocity and inversely proportional to the acceleration due to gravity. Understanding this relationship is crucial in physics, especially in kinematics.